1. The diagram below shows a portion of a
DNA molecule. Show with the help of
diagrams how this portion replicates.
2. Which of the following steps in DNA
replication involves the formation of new
a) Initiation of an origin of replication
b) Elongation by a DNA polymerase
c) Unwinding of the double helix
3 . The difference in leading- versus laggingstrand synthesis is a consequence of –
a) the antiparallel configuration of DNA
b) DNA polymerase III synthesizing only in
the 5’-to-3’ direction
c) The activity of DNA ligase
d) Both (a) and (b)
4. Okazaki fragments are—
a) synthesized in the 3’-to-5’ direction
b) found in the lagging strand
c) found in the leading strand
d) made of RNA
5. The replication of a DNA molecule occurs in
a series of coordinated steps requiring the
action of several enzymes.
a) What is DNA replication?
The process by which new copies of DNA are
b) DNA replication is termed semiconservative replication. What does
this imply about the process? (1 mark)
The newly formed DNA consists of an
original strand and a synthesised one.
c) One of the first steps in DNA replication
involves the formation of a replication fork.
Name the enzyme responsible for the formation
of a replication fork and briefly describe how
this enzyme modifies a double helix to form this
Helicase is the enzyme. It breaks the hydrogen
bonds between the complementary bases.
d) What are the roles of DNA polymerase in
the replication process? (2 marks)
Catalysis the polymerisation of
deoxyribonucleotides alongside a DNA
strand which it “reads” and uses as a
Plays a role in proofreading.
Removes RNA primers and replaces them
e) The lagging strand of the replication
fork is synthesized in short fragments
known as Okazaki Fragments. Suggest
a reason for the synthesis of these
Since DNA polymerase III can lay down
new nucleotides only in the 5’ to the 3’
f) What is the role of ligase in the formation of
the lagging strand?
Ligase joins the Okazaki fragments together by
helping the formation of phosphodiester
g) During which part of the cell cycle is DNA
replication likely to occur? (1 mark)
During S phase of interphase.
THE GENETIC CODE
6. The following are mRNA codons for three
amino acids and a stop codon.
Serine UCA Lysine AAA
a) Give an appropriate nucleotide base sequence
i) the DNA strand complementary to an
mRNA codon for serine;
mRNA codon = UCA and UCG
DNA codon = AGT and AGC, respectively
Serine UCA Lysine AAA
a) Give an appropriate nucleotide base sequence for:
an anticodon for glutamine.
mRNA codon = GAA and GAG
anticodon = CUU and CUC, respectively
b) Giving examples from the codons above, explain why
the third base of a codon is less critical than the first
The first two bases determine the amino acid e.g. GAA
and GAG both code for glutamine.
7. The flow diagram shows the course of an
experiment with a virus and the bacterium
a) Which molecule in the virus was being
i) radioactive sulphur, 35S
ii) radioactive phosphorus, 32P?
Phosphate group of nucleotides.
b) Explain how the viruses collected from
culture A became labelled.
Viruses use labelled amino acids to build
c) Explain the difference in radioactivity of the
viruses at the end of the experiment.
Radioactive sulphur: No radioactive viruses
were collected as the protein coat did not enter
Radioactive phosphorus: Some radioactive
viruses were collected as they injected their
DNA into bacteria.
d) Suggest why 14C was not used to label the
viruses in this experiment.
As carbon is present in both amino acids and
nucleotides. It would not be possible to trace
proteins and nucleic acids if radioactive carbon
is used as this is present in both.
e) This classical experiment was carried out by
Hershey and Chase in1952. What important
conclusion did they come to, regarding phage
genetic material, following the results of this
experiment? (1 mark)
DNA codes for the protein coat.
8. Cells of the bacterium E. coli were grown for many
generations on a medium containing the heavy isotope
of nitrogen, 15N. This labelled the entire DNA in the
The cells were transferred to a medium containing 14N and
allowed to grow. During each generation of bacteria, the
DNA replicates once. Samples of the bacteria were removed
from the culture after one generation time and after two
generation times. The DNA from each sample was extracted
and centrifuged. As the DNA containing 15N is slightly
heavier than that containing 14N, the relative amounts of
DNA labelled with 14N and 15N can be determined.
The diagram below shows two reference tubes
and the results of this experiment.
a) Which part of the DNA molecule in the
original culture would have been labelled
with 15N? (1 mark)
b) Explain why the DNA occupies an
intermediate position after one generation
in the 14N containing medium.
Weight of DNA is intermediate between
heavy and light types. It is made of one
heavy and one light strand. (2 marks)
c) Complete the diagram to show the
position of the band or bands of DNA
after two generations in the 14Ncontaining medium (Tube X). (1 mark)
d) This neat experiment was designed by
Matthew Meselson and Franklin Stahl
in 1958 in order to evaluated 3 thencurrent models of DNA replication.
Which of these models for DNA
replication is supported by the results
of this experiment and is still
9 Read through the following account of DNA and
protein synthesis, and then write on the dotted
lines the most appropriate word or words to
complete the account.
A molecule of DNA (deoxyribonucleic acid) consists
of two polynucleotide strands with complementary
pairs of organic bases. Each nucleotide consists of
a sugar called DEOXYRIBOSE, a PHOSPHATE group
and an organic base. Pairs of bases are held
together by weak HYDROGEN bonds. During the
first stage of protein synthesis the DNA strands are
separated and a complementary molecule of mRNA
is synthesised in the nucleus using one strand of
DNA as a TEMPLATE .
Where the base cytosine occurs in the
DNA strand, the newly synthesised
molecule contains the base GUANINE
and where the DNA has adenine, the
newly synthesised molecule has URACIL.
The newly synthesised molecule then
passes into the cytoplasm and becomes
attached to a RIBOSOME where
synthesis of the polypeptide chain
Complete the diagram below by adding
the corresponding nucleotide bases to:
the DNA strand;
the tRNA molecules to form the
appropriate anticodons for the amino
acids X, Y and Z.
b) The following are mRNA codons for three
Identify amino acid Z.
Which of the amino acids X, Y and Z has or have a
degenerate code? Explain your answer.
X and Z – different codons code the same amino acid.
c) What is the function of the base sequence
It determines the sequence of amino acids in a
polypeptide. Every three bases make up a
codon and code for one amino acid. A start
codon (AUG) denotes where the polypeptide
formation starts while a stop codon (UGA, UAA
d) Describe the part played by ribosomes
in protein synthesis. (2 marks)
The ribosome binds mRNA and tRNA
molecules. rRNA in the large subunit acts as an
enzyme that links amino acids together as
peptide bonds form.
e) Describe the role of transfer RNA in
protein synthesis. (3 marks)
A tRNA molecule binds an amino acid
in the cytoplasm and delivers it to the
ribosome. Its anticodon is
complementary to the mRNA codon.
11 a) A piece of mRNA is 660 nucleotides long
but the DNA non-template strand from which it
was transcribed is 870 nucleotides long.
i) Explain this difference in the number of
nucleotides. (1 mark)
DNA is longer
due to introns.
11 a) A piece of mRNA is 660 nucleotides long
but the DNA template strand from which it was
transcribed is 870 nucleotides long.
ii) What is the maximum number of amino
acids in the protein translated from this
piece of mRNA? Explain your answer.
Due to the triplet code: 660/3 = 220
Max No. of amino acids: 220-1 (due to a stop
codon that does not code for an amino acid)
b) Complete the table to give two
differences between the structure of
mRNA and the structure of tRNA.
1. Has codons
2. Linear molecule Molecule is folded
WORK OUT Nos. 18, 19
[about lac operon] for
Friday 7 March 2014
CONTROL OF GENE EXPRESSION IN
Control of gene expression can occur at which of the following
Splicing of pre-mRNA into mature mRNA
Initiation of translation
Initiation of transcription
All of the above
15 Negative control of transcription in a prokaryotic cell involves
__________molecules that alter the conformation of _______________
proteins that bind to DNA and preventtranscription.
a) operator; repressor
b) activator; RNA polymerase
c) activator; operator
d) effector; repressor.
What is an operon?
A region of DNA involved in regulation of transcription
A cluster of genes that are expressed as a single unit
A DNA-binding motif
A regulatory protein that enhances transcription
17 What effect would the presence of lactose have on a lac
a) The repressor would bind to the operator site of the
b) Lactose will bind to the operator site of the operon
c) The lac operon would be transcribed.
d) It would have no effect.
18 The figure below shows a
diagrammatic representation of the
lac operon including the Promoter
region, Operator region and three
a) What is an operon? (2 marks)
An operon is a segment of DNA containing
adjacent genes including structural genes, an
operator gene, and a regulatory gene. It is a
functional unit of transcription and genetic
b) Briefly describe the function of:
i) The promoter region;
The DNA region, usually upstream to the
coding sequence of a gene or operon, which
binds and directs RNA polymerase to the
correct transcriptional start site and thus
permits the initiation of transcription.
The operator region;
A gene that activates the production of
messenger RNA [transcription] using
adjacent structural genes. The repressor
protein binds to the operator region.
iii) The three structural genes lacZ, lacY and
These are three structural genes coding
for three different proteins – beta
galactosidase, galactosidase permease
and galactosidase transacetylase.
c) List ONE way in which gene
expression in prokaryotes differs from
gene expression in most eukaryotes.
Eukaryotes have a nuclear envelope,
which prevents simultaneous
transcription and translation.
Gene expression involves a
polycistronic mRNA but a
monocistronic one in eukaryotes.
19. Colonies of the bacterium Escherichia coli
grow rapidly on culture media based on glucose.
When transferred to a culture medium based on
lactose, growth rates of the colonies slow down
temporarily but subsequently reach the same
rates characteristic of glucose-based media.
Analysis of the colonies growing on lactosebased media indicated the presence of –
galactosidase, an enzyme that hydrolyses
lactose to glucose. This enzyme is not present in
colonies grown on glucose-based media.
a) –galactosidase is an inducible enzyme. What
is an inducible enzyme? (2)
Produced only in the presence of the inducer
molecule, i.e. lactose.
b) What is the principal difference
between inducible enzymes and
constitutive enzymes? (2)
Inducible enzymes are made only in the
presence of their inducer molecules,
while constitutive enzymes are made all
c) Briefly explain why –galactosidase is not
synthesised when E. coli is cultured on media
lacking lactose. (5)
Repressor molecules are made all the time in
In the absence of lactose, the repressor molecule
is active and binds to the operator region on the
The binding of the repressor to the DNA,
prevents RNA polymerase from binding to the
This means that transcription cannot occur. If
mRNA is absent, then no translation occurs and
enzymes are not made.
d) Some E. coli will synthesise –
galactosidase in the absence of lactose.
Suggest an explanation for this
A mutation in the gene results in a
different shape of the repressor molecule
which cannot bind to the operator , thus
RNA polymerase can bind to its promoter
and transcription proceed. Translation
A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later.